 In this third lecture of module 6 on mass transfer across interfaces, today we continue the discussion on resistances in the liquid phase and at interface, their importance in certain applications and observed effects and then we develop on the theory of mass transfer bit. While we are talking about the evaporation, when the liquid phase resistance was negligible, we had minus 1 by a d w by d t giving you rate of loss of weight of the drop of liquid per area related to the driving force which is p by r t that would be the concentration in the vapor corresponding to the saturation vapor pressure and the resistance which would be only in the liquid phase that would be the radius a by diffusivity, radius only in the gas phase that is the radius a of the drop by diffusivity of the vapor in the gas phase. If the resistance in the liquid where to come into picture, it would be added to this gas phase resistance. The presence of m here is only because of weight on the left hand side. So, it turns out that if we have very small drops, then the rate of evaporation corresponds to that in vacuum and that ties in with your visualization that in the final stages, the drop would tend to just flash off into the surrounding gas. However, for different drop sizes, let us say for about 10 microns, the correction because of this liquid phase resistance could be measurable about 5 percent, but when the drop becomes about a tenth of a micron, then the resistance in the liquid will cause the increase in the lifetime of the drop by a factor as high as 6, which means since these final stages when the radius of the evaporating drop becomes very small, this gas phase resistance has reduced so much that the liquid phase resistance can now start controlling the magnitudes of the lifetimes. We will see that the lifetimes of droplets, fine droplets can be of significance in certain effects and some applications that we will see near the end of this lecture. While talking about the evaporation, a factor which we are not mentioned yet is the cooling of small drops. As evaporation occurs, the latent heat will be taken by the evaporating liquid from itself and therefore, that effect can be significant for certain liquids like toluene than for let us say water. It is here that we now begin thinking about the importance of the interfacial resistance R i related to the gas phase resistance R g and how it would determine the rate of evaporation of small drops. We would find in certain situations that the rate of evaporation becomes a sensitive function of the traces of impurities present. Rate of evaporation depends significantly on minute quantities of impurities that would clearly indicate that we have now interfacial resistance coming into picture. For instance, we had looked at these numbers earlier a monolayer of a long chain molecule of alcohol or acid at room temperature could cause an increase in the interfacial resistance from about 2000th of a second per centimeter to about 10 seconds per centimeter, which means it should be able to retard the rate of evaporation by a factor 10 by 0.002 that is about 5000 times. So, presence of these long chain alcohols or acids could actually increase the lifetime of the droplets by over 3 orders of magnitude. This will be of significance and of interest in couple of applications you could think of let us say coal mines. I will spell out the exact application little later and an effect which I will suggest as the presence of fog or rather smog in the winter months in cities like Delhi. Lifetimes of fine water droplets in dry air when the drop size is about let us say about a micron radius, it could be increased from a few milliseconds to almost a minute just by the effect of the interfacial resistance imparted by presence of a monolayer. Now, all it would take is very slight level of impurities present to form a monolayer and you could witness a dramatic change in the time it takes for evaporating such fine droplets. Take another example, if air has about 80 percent relative humidity then the life of a 10 microns drop will increase from about 2.5 seconds to 1300 seconds. Monolayers are not alone able to increase the lifetime of such droplets. Presence of dust, non volatile impurities or other products like of oxidation and decomposition could also retard the rate of evaporation especially if the drops are very small. In context of smog, if tarry material is present that would be responsible for delaying dispersion of the smog under the conditions of isothermal distillation that is the effect that you could think of the smog is harder to clear rather than the fog which would appear in cleaner air. So, more polluted the atmosphere harder will be the problem of smog and associated health risk because it is a smog which you inhale and the particulates with the adsorbed impurities could be a major problem for people sensitive to these impurities from respiratory perspective. If you take mercury you all know that mercury is volatile, mercury is also poisonous. So, it is a common advice and I think almost unwritten rule in laboratories that if you spill mercury never use a vacuum pump vacuum pump to clean up the spilt mercury. If you apply suction will cause mercury to evaporate and that is dangerous. So, you must never do that. So, the evaporation of mercury also could be retarded by presence of skin of oxidized products. This may remind you of the picture of a well known model for reaction of gas solid kind. The oxidation product here offers similar rule for a distance as the ash layer would find you would find is offering in that model. And for mercury droplets the evaporation could be slowed down by almost three orders of magnitude. Now, you can think of a slightly different scenario. If the drop that you are considering evaporation of is present in a turbulent stream of air. Unlike what we have seen so far the gas phase resistance was given by A by D. The radius of the droplet by the diffusivity in the gas phase that is true only if the air around that drop is stationary stagnant. If the air is present in turbulent conditions one would expect the hydrodynamic factors to come into picture. The velocity density viscosity and drop size all would come into picture that will be all combined in a dimensionless number Reynolds number. And we expect the physical properties also to come into picture. The physical properties like viscosity and diffusivity. So, resistance under turbulent conditions will be lesser or the evaporation rate will have to become higher if you are turbulent conditions. You can visualize this slightly differently by looking at a single drop if I have a drop of liquid and air around it. If this air is stagnant then in principle the vapor concentration the liquid evaporates at the surface and then diffuses into the bulk of air. If air is stagnant this concentration drop would continue till infinity right. However, and that would mean that in such a case the hypothetical film of film theory will be infinitely thick. In principle the diffusion can occur till infinity. One might say an approximation that is done in film theory as whatever is the gradient of concentration you persist with the initial gradient and see where the concentration becomes equal to the bulk concentration. And then you identify that as the thickness which is the thickness of the film in the film theory that delta right. If you for instance now replace this stagnant air by air which is stirred stirred to have random or chaotic motion turbulent motion then what it means is that you will be able to destroy the concentration gradient in the bulk. And the concentration profile will become a uniform concentration profile now to a closer proximity to the drop. Our new film thickness will be lesser under turbulent conditions. Let us say we call it delta t greater the intensity of this mixing or greater the intensity of turbulence here thinner will be this stagnant film of film theory delta t will get smaller right. Which means the resistance which is attributable to this gas film will get reduced as you increase the stirring resistance gets reduced therefore the rate of operation will be higher. Now that is common right it is analogous but that example of dissolving sugar in a cup of t by stirring is an analogous example left to itself sugar will dissolve in a cup of t but it will take a very long time you will not be able to reach your classes after breakfast. But in order to make sure that that time of dissolution is manageable very quick you stir it and then that stirring is equivalent to the stirring of gas phase. There is dissolution of sugar particles here it is the evaporation of this droplet greater the extent of stirring greater will be the rate of evaporation. So, the rate of evaporation that we have talked about so far minus 1 upon a d w by d t that expression in terms of the vapor pressure p and diffusivity d that will have to be multiplied by a certain factor that factor should involve a major of the stirring which is basically reflected in Reynolds number and certain properties. So, our expression for rate of evaporation will now have on the right hand side a certain correction term that correction factor some of you might be able to see should be as 1 plus b times Reynolds to the power half. Now, what is this factor as I was arguing the for the turbulent conditions characterized by Reynolds number we will have an increase in the rate of evaporation or decrease in the resistance and increase in the rate of evaporation will be brought in through this Reynolds to the power half term, but we also mentioned a constant b here this constant b you would be able to see will depend on properties relevant to mass transfer that would include the diffusion cohesion and the kinematic viscosity of gas or if you have to think in mass transfer terms some of you might be able to recollect that b must contain the dimensionless properties group the Schmidt number. In particular b will come out to be 0.6 Schmidt to the power 1 by 3 in a wide range of conditions this correlation holds well that comes from Rans and Marshall correlation or Frossling equation. So, under turbulent conditions we realize that the gas phase resistance will go down and the hydrodynamics will come into picture and the physical properties will come into picture and this correction factor has been shown to be in good agreement with experiment in context of systems like particulate clouds dust smokes and mists. Now, one application that I would like you to think of is say of coal mines think about how droplets of water and their evaporation might be of significance and of some utility in a coal mine. Think of what potential problems you might have in a coal mine and think of what might possibly be application for water droplets and for controlling their evaporation. So, the coal should be completely dry in order to increase its calorific value. So, basically the presence of water or moisture is undesirable. So, basically we can utilize this fact because under pressure inside a coal mine is lower the atmospheric pressure is like we were. So, basically that will help in reading the evaporation of water and this will be more of removal of moisture from the coal. Good attempt think more let us say we are thinking of problems imagine for an instance that you are a worker in a coal mine that will help. What problems will you face there? You might be equipped you might be equipped to everything and yet you might want to eliminate the difficulties. As this mining process is going on you are bound to create very very small coal particles. They will fly around right. You would want them to be all also settle and captured. So, what is done here is that you spray water find droplets of water which are able to coalesce these particles and then keep them together in agglomerated form. Now, the difficulty is that because you would be atomizing water into very fine droplets. If left to the purity of water itself droplets of water will be generated about 1 micron size or there about will tend to evaporate very rapidly. So, they would not be able to bind the coal particles together. You want water to be present there forming bridges between the particles. So, that the particles can be kept together and that can be done by retarding the rate of evaporation. For instance here while mining water content is not an issue eventually of course you want as minimum water as present as possible. But while mining you want the ambient air to be as free of particles as possible. So, you actually try to retard the rate of evaporation. You could do it by adding some monolayer forming materials. Once again the satil alcohol comes into picture. Traces of satil alcohol in water which is previously dispersed in it will be able to prevent the evaporation of these aerosol droplets and you would be able to bind the coal dust together. So, droplets of over 12 micron radius would be rendered relatively stable although without this satil alcohol or the monolayer the complete evaporation would occur in just a few seconds. Now, with that we move on to the solute transfer at gas liquid surface. I had broached this topic while we are at the stage where we added to that gas phase resistance A by D R L in the equation giving you the rate of evaporation in terms of weight loss per time per area as an example moving away from evaporation. Now, we are really talking about the same situation and I had mentioned there that we have the gas phase here the liquid phase is here. If we had pure gas there would still be some resistance for arrival of the pure gas molecules to the interface before they could be transferred into the liquid. If we have pure gas and pure liquid in the first instance then the concentration at the interface will be C A star and it will drop to 0 within no distance that will mean the gradient is infinite flux is infinite. So, liquid phase resistance at 0th instance is 0 entire resistance is in the gas phase that gas phase resistance is minimum if the gas is pure. If it were a dilute gas A like solute A in inert I then that gas phase resistance will be even greater. The least resistance you can have is when there is no inert and you have only the pure solute in the gas phase. Then I argued that as transfer occurs that concentration at the interface will remain saturation concentration where the gas is pure here and this gradient now diminishes becomes finite and lesser and lesser as time progresses. That corresponds to concentration of diffusing A advancing deeper into the liquid as more and more liquid thickness comes into picture the liquid phase resistance increases. And I had argued that even for pure gas and pure liquid case the time over which this gas phase resistance becomes is significant is only very short within about 10 picoseconds the resistance in the liquid becomes so large probably about 9 times the resistance in the gas phase. So, that overwhelming or the predominant resistance becomes the resistance resistance in the liquid phase. So, what it means is that for pure gas systems except for that very small fraction of a second 10 picoseconds of almost order of 10 raise to minus 11 seconds the gas phase resistance becomes a factor not in the picture. So, for all practical applications where contact times are of the order of a few seconds at the list may be 10s and 100s of seconds in some cases that initial flicker is not important. That initial moment where the gas phase resistance is important is not of significance in practice. So, if you have pure gas we could safely say that gas phase resistance can be neglected is a liquid phase resistance which can come into picture may be the inter facial resistance. So, with that background let us move on here we are talking of absorption of such gases as carbon dioxide or ammonia in water. And for all practical purposes the take home message should be R L is important very important in controlling the rate of absorption. And this is actually true also for desorption desorption actually just the reverse of absorption only the direction of transfer is opposite. There are of course, theoretically there are certain other differences will probably able to talk about those later, but main point here is that both for absorption and desorption where the direction of solute transfer is opposite the liquid phase resistance is still the dominant and controlling resistance. And how large is this resistance typically 100 to about 1000 seconds per centimeter. What value resistance of liquid will have will depend of course, on the conditions which are of the two kinds just now we discuss those. The hydrodynamics which will bring in the Reynolds number as dimensionless group and other physical properties. So, Reynolds and other physical properties Reynolds number and other physical properties would become factors on which the liquid phase resistance will depend. Now to initiate some discussion of theory and models of mass transfer for the whole class in general. We may start with this simplicity that the liquid phase is so great in extent that it can be taken practically as infinite. Infinite is hard to imagine, but mathematically it turns out to be a to be an expedient which simplifies the mathematics. It might appear counterintuitive at first instance, but this is true. It is easier to analyze transfer in infinite liquids. So, if we have a situation like the one I just have shown that interface is regarded as planar on one side where the gas phase on the other the liquid phase. And we can regard the entire transport to be one dimensional problem. And for generality look at a transient situation unsteady state situation where time has a role to play. Then the solute concentration which in this equation is capital C will depend on time and how far that point under consideration is from the interface where the transfer begins. At the interface like in that case with the concentration equal to C A star, we have a constant concentration there. A boundary condition is corresponding to a constant concentration. As we go away from that interface into the bulk of liquid the concentration will decrease because the gas is being transferred from the gas phase to the liquid. So, the gas first dissolves into the liquid at the interface the concentration is raised to the highest value. And then in the bulk of the liquid the concentration diminishes. How at any time the concentration will be dependent on x or at any x how the concentration will increase with time will be governed by this equation which is nothing but fixed second law or the diffusion equation. So we have the transient term dou C by dou t equal to the diffusion cohesion in the liquid of the solute and the second partial of the concentration with the space coordinate x. x is here the distance measured from the interface into the depth of liquid. Potentially x starting from value of 0 could increase to infinity provided we can regard the liquid to be infinite. It is here that I want to make a few comments for complete clarity. Obviously you would argue that there is nothing infinite in the universe. The universe itself is finite. So how in chemical engineering practice could we justify an infinity? No depth of liquid, no extent of spatial dimension of a vessel can really strictly be infinity. So it is here that we come to accept this that most of the chemical engineering practice involves comparison of scales. So it is comparison of the length scale relative to certain characteristic length scale that leads to this approximation. If the depth of liquid is large compared to let us say a typical depth of penetration corresponding to a given time of contact you can regard a few millimeters of liquid also to be infinite, a few centimeters definitely. So this infinity is not a mathematical infinity but a physically justifiable approximation to infinity and it simplifies our analysis quite a lot. I am not going into the details of the solution of this equation but suffices to say that we need one initial condition and two boundary conditions. If we take this example of pure gas and contact with the liquid one boundary condition will correspond to x equal to 0 the concentration is C A star and the other boundary condition will be when x tends to infinity far away from interface the concentration will be 0 because we had a pure liquid and at time 0 everywhere for x greater than 0 the concentration is 0 and then there is that understanding that at time 0 at x equal to 0 the concentration is instantly equal to C A star. Practically it will take some time for the surface concentration to rise from 0 to the C A star but that takes time of the order of a few milliseconds as has been proven by researchers through actual experimental measurements but that is a very good approximation. We can regard the interface to be immediately saturated and therefore we will be able to postulate the initial and boundary conditions then the problem becomes well posed and could be solved. Fortunately for this infinite extent case the solution of this equation is possible analytically you can use either separation of variables or you can use Laplace transforms because this is a linear problem with constant cohesion provided diffusion cohesion is regarded as constant it is possible to solve this and you get an error function complementary solution. Once you have that then evaluation of gradient and then flux will give you finally the rate of absorption in this form d cube by dt is equal to the total rate of absorption will be equal to total area and this concentration driving for C s minus C infinity where C s please note that C is the concentration of dissolved gas this gas here is to be understood as dissolved gas. So, C represents the concentration in the liquid that is the dissolved gas component A in liquid C s is corresponding to the saturation what I indicated C s star is C s here C infinity is the concentration at infinity in our other case it would have been 0 and then we get a factor square root of d by pi t d is diffusivity t is time. If you compare this with what we had been talking of earlier this is your permeability cohesion or mass transquestion or is reciprocal will be the resistance in the liquid phase right. So, we can immediately identify the resistance in the liquid phase now dependent on time after all as time progresses the depth of penetration of the solute is increasing therefore, the resistance is increasing that is what is shown by this the mass transquestion K L which is reciprocal of R L is root of d by pi t or it shows that R L increases as time increases dependence is square root time. So, if we take some representative values for diffusivity for many solutes small solute molecules in liquids like water practically all liquids order of magnitude for diffusion coefficient is about 10 to the power minus 5 centimeter square a second and with the time of exposure of the order of a second we will find that resistance is about 500 seconds per centimeter at given you this value as one of the representative values for liquid phase resistance. However, the time of exposure if it is as large as 100 seconds 100 fold then as expected from this dependence R L proportional root t there is a 10 fold increase in this resistance. So, 500 becomes 5000 seconds per centimeter 100 fold increase in the exposure time will increase the resistance from 500 to 5000 seconds per centimeter. So, can we say we have applied penetration exactly yeah this is basically one may say penetration theory provided this time can be regarded as constant, but the safer statement will be that K L proportional to 2 square root of d means that we have use some kind of surface renewal theory. Penetration model becomes a special case in this category when the lifetime becomes constant in a general surface renewal theory we will have a distribution of lifetimes of these elements or eddies from 0 to infinity. But for any surface renewal model where the surface is renewed by fresh liquid after certain intervals or whatever distribution the dependence will come out to be same K L proportional to root d and this is the main difference between these renewal models dependence of K L on diffusivity which is different from the other models in film model you should expect K L proportional to d and in the boundary layer theory applications we will have diffusivity to the power 2 by 3 those are typical other important models and in experiment this is the best agreement because it is somewhere in between the whole range of exponents we see on diffusivity 0 to about 0.8 or 0.9. For desorption what would be the differences you can think quickly here if it is desorption you can think of now transfer beginning at infinity in the bulk of liquid and for simplicity we may say the concentration in the surface could be kept at 0. It could be if you have a rapid flow of an inert gas at the interface such that the moment solute molecule is released by the liquid it is carried away and concentration of that solute in the gas phase drops to 0 or alternatively we could say that C s is equal to 0 by actual application of vacuum the moment a solute molecule is released it just flashes off and vanishes. So, if we take C s equal to 0 then we could use the same equation for desorption and another quantity which might be of interest besides the rate of absorption is how much of the total solute gas we have absorbed how much of that solute a has been absorbed or desorbed that could be obtained by integrating this previous equation we need q. So, if we integrate you can see here t to the power minus half when you integrate that you get t to the power half divided by 2 divided by half which means t to the power half which will be root d t by pi and divided by half means 2 here. So, q will become equal to 2 a C s minus a infinity square root d t by pi this is what the result is of integration total moles transferred in time t is given here. Now, the question which I have sort of passed very quickly I just mentioned that presuming the liquid to be infinite helps in analysis what if the liquid cannot be regarded as infinite you might think of situations like a foam film or a thin film of liquid into which absorption is occurring or from which desorption is occurring. Now, you might have only a few tens of microns thick film about 50 microns maybe 100 microns thick film and if now absorption is occurring in the film from both faces the concentration profile of the diffusing solute from either side will be able to meet somewhere in between. So, the finite extent of the film comes into picture. First there are possibilities of transfer occurring from both directions. Secondly, because of finite extent of the film the concentrations profiles are going to collide. So, one phase starts seeing the effect of the other phase. So, the finite size effect comes into picture those effects come into picture we do not have that kind of situation in an infinite medium. Even if a solute manages to reach infinite depth of liquid it will take infinite time we will not never be able to see what happens at the other boundary. What it means is that now the analysis becomes that much harder you would see that the techniques like separation of variables or Laplace transforms fail. Those are only of utility in linear constant cohesion partial differential equations pertaining to infinite medium. The moment you have finite medium those things do not work. Method of combination of variables you remember similar transformation also will not work because that is only true for infinite medium. We cannot have the combination of variables in that sense possible. So, at times if the problem is linear constant cohesion you might still see Laplace transform being useful, but if the problem is non-linear if there is a reaction which is non-linear then Laplace transformation would not work. Separation of variables can work with constant cohesion. If the diffusion cohesion becomes dependent on concentration then there is no way we will be able to solve that problem. In such cases then you have no alternative but to go for numerical solutions. And obviously today we have lots of numerical techniques available and software packages available which could be judiciously used to solve such problems. But then once you get into a computational solution the direct insight into structure of problem is lost because you are not able to see the analytical dependence of let us say rate of absorption or of mass transfer question are equivalent on the individual parameters. Then we can think of some situation like absorption or desorption in a stirred system. If you have a tank containing liquid and bubbles which are supplying gas to the liquid or which are picking up solute from the liquid super saturated solution performing desorption we would expect now the transfer to be depend on mechanical factors rate of agitation. How agitation is caused? They might be changes in density or temperature depending on what is the cause of stirring how do you cause the mixing or agitation. All these complications will mean that now for all practical purposes you may not be able to deduce the liquid phase resistance from first principles. Each individual case will have to be dealt with separately. I will just mention this one last point another complicating factor is one corresponding to surface turbulence. I have mentioned some time back the notion of surface turbulence. Imagine that in the surface of a pure liquid if you measure surface tension it should be same surface pressure is 0. If it is pure liquid there is no surface pressure corresponding to impurities. But if you have a solution then the situation can be different. It can happen that a solution like of ether in water might exhibit some heterogeneity of surface concentration. Momentarily there are different concentrations on the surface that means there are different surface pressures and therefore there are these unsteady surface pressure gradients. If there are surface pressure gradients or surface tension gradients there will be flow in the surface. If there is a flow in the surface then by virtue of the finite viscosity of liquid the liquid underneath will be dragged. So because of presence of non-uniform surface concentration they can be unsteady movement in the surface leading to mixing in the adjacent liquid layers and therefore it can compound effect on mass transfer. I think I do not have a time enough time here to talk about surface turbulence but I will just end this lecture with one simple counterintuitive example and I may take you down to a proverbial space. You all are aware of a proverb adding oil to fire. We expect adding oil to fire will increase the extent of the destructive nature of fire. Can we have some exception here? I will give an example. Let us say we have a saturated solution of ether in water that will contain about 5.5 percent of ether in water. Now there are only these two components ether and water and you light a mastic and bring it close to the surface of this solution. What do you expect will happen? Mostly water little ether just about 5.5 percent. You bring a mastic close to the surface of this solution. What would happen? Ether is implemable. Ether is soluble to the extent about 5.5 percent that saturation at the surface will catch fire. Now I have to tell you something. Maybe I can ask you another question. These are questions we do not think in terms of normally although a little reflection will tell you that these are common sense questions. Question is do liquids burn? Yes. Correct? Liquids do not burn. It is liquids which give out vapors and then vapors burn. The same thing is happening here. There is ether dissolved in water. If ether can evaporate it can catch fire. So you are right. Near the surface the ether vapor will catch fire. How exactly this happens? We may think about later and the surface turbulence will come later. But I will just end this lecture with one more question and you have to answer that. Now I take the same ether solution in water 5.5 percent, weight percent and I add just the drop of oleic acid on top of this surface. I do the same thing. I bring a lighted mastic close to the surface. What would happen now? You will take in the separation of ether. Ether and therefore? It will not catch fire. It will not catch fire at all. Yes, it is not that inflammable as ether. So you really cannot now light up the vapor. What you can do is, you can check this hypothesis. Take a glass rod and whip the surface a little bit, disturb it. You would be able to temporarily overcome the effect of oleic acid. There will be fire. So here, oleic acid which is otherwise possible to burn can extinguish fire. I think that answers one of the questions which was raised last time. Can you reduce the evaporation of petroleum products or volatile products? Yes, using suitable monolayers, you can increase the interfacial resistance and thereby suppress the vaporization and therefore, the inflammability. If you have water on top of it and micro emulsion kind of fuels, of course, you are a lot safer. We will stop here for today.